`int_0^(pi//2) x^2 cos 2x dx`

Answer

To solve the integral \( I = \int_0^{\frac{\pi}{2}} x^2 \cos(2x) \, dx \), we will use the property of definite integrals and integration by parts.

### Step 1: Use the property of definite integrals
We know that:
\[ \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \]
For our case, let \( a = \frac{\pi}{2} \). Thus, we can write:
\[ I = \int_0^{\frac{\pi}{2}} \left(\frac{\pi}{2} - x\right)^2 \cos\left(2\left(\frac{\pi}{2} - x\right)\right) \, dx \]

### Step 2: Simplify the expression
Now, we simplify the expression:
\[ \left(\frac{\pi}{2} - x\right)^2 = \frac{\pi^2}{4} - \pi x + x^2 \]
And,
\[ \cos\left(2\left(\frac{\pi}{2} - x\right)\right) = \cos(\pi - 2x) = -\cos(2x) \]
Thus, we can write:
\[ I = \int_0^{\frac{\pi}{2}} \left(\frac{\pi^2}{4} - \pi x + x^2\right)(-\cos(2x)) \, dx \]
This becomes:
\[ I = -\int_0^{\frac{\pi}{2}} \left(\frac{\pi^2}{4} - \pi x + x^2\right) \cos(2x) \, dx \]

### Step 3: Combine the integrals
Now we have two expressions for \( I \):
1. \( I = \int_0^{\frac{\pi}{2}} x^2 \cos(2x) \, dx \) (original)
2. \( I = -\int_0^{\frac{\pi}{2}} \left(\frac{\pi^2}{4} - \pi x + x^2\right) \cos(2x) \, dx \)

Adding these two equations gives:
\[ 2I = \int_0^{\frac{\pi}{2}} x^2 \cos(2x) \, dx - \left(\frac{\pi^2}{4} \int_0^{\frac{\pi}{2}} \cos(2x) \, dx - \pi \int_0^{\frac{\pi}{2}} x \cos(2x) \, dx + \int_0^{\frac{\pi}{2}} x^2 \cos(2x) \, dx\right) \]

### Step 4: Evaluate the integrals
Now we need to evaluate the integrals:
1. \( \int_0^{\frac{\pi}{2}} \cos(2x) \, dx = \left[\frac{1}{2} \sin(2x)\right]_0^{\frac{\pi}{2}} = \frac{1}{2} \)
2. Let \( J = \int_0^{\frac{\pi}{2}} x \cos(2x) \, dx \)

### Step 5: Use integration by parts for \( J \)
Let \( u = x \) and \( dv = \cos(2x) \, dx \). Then,
\[ du = dx, \quad v = \frac{1}{2} \sin(2x) \]
Using integration by parts:
\[ J = \left[ x \cdot \frac{1}{2} \sin(2x) \right]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} \frac{1}{2} \sin(2x) \, dx \]
The first term evaluates to \( 0 \) and the second term evaluates to:
\[ -\frac{1}{2} \left[-\frac{1}{2} \cos(2x)\right]_0^{\frac{\pi}{2}} = -\frac{1}{2} \left(-\frac{1}{2} - 1\right) = \frac{3}{4} \]
Thus, \( J = \frac{3}{4} \).

### Step 6: Substitute back into the equation for \( I \)
Now substituting back:
\[ 2I = I - \left(\frac{\pi^2}{4} \cdot \frac{1}{2} - \pi \cdot \frac{3}{4} + I\right) \]
This simplifies to:
\[ 2I = I - \left(\frac{\pi^2}{8} - \frac{3\pi}{4} + I\right) \]
Rearranging gives:
\[ I = \frac{\pi^2}{8} - \frac{3\pi}{4} \]
Thus, the final answer is:
\[ I = \frac{\pi^2}{8} - \frac{3\pi}{4} \]